ГДЗ по геометрии 7 класс Атанасян Задание 512

\[\boxed{\mathbf{512}\mathbf{.}\mathbf{ОК\ ГДЗ - домашка\ на}\ 5}\]

\[Рисунок\ по\ условию\ задачи:\]

\[\mathbf{Дано:}\]

\[\text{ABCD} - трапеция;\]

\[\text{BC} = a;\text{AD} = b;\]

\[\text{MN} \parallel \text{AD};\]

\[S_{\text{AMND}} = S_{\text{MBCN}}.\]

\[\mathbf{Найти:}\]

\[\text{MN} - ?\]

\[\mathbf{Решение.}\]

\[1)\ \text{MN} = x.\]

\[(b + x)\text{BH} = (a + b + 2x)\text{BF}.\]

\[4)\ S_{\text{ABCD}} = S_{\text{AMND}} + S_{\text{MBCN}}\]

\[\frac{1}{2}(a + b)\text{BH} =\]

\[= \frac{1}{2}(b + x)\left( \text{BH} - \text{BF} \right) + \frac{1}{2}(a + x)\text{FB}\]

\[(a + b)\text{BH} =\]

\[= (b + x)\left( \text{BH} - \text{BF} \right) + (a + x)\text{BF};\]

\[\text{aBH} - \text{xBH} = \text{aBF} - \text{bBF};\]

\[(a - x)\text{BH} = (a - b)\text{BF}.\]

\[5)\ \left\{ \begin{matrix} (b + x)\text{BH} = (a + b + 2x)\text{BF} \\ (a - x)\text{BH} = (a - b)\text{BF}\text{\ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \text{BH} = \frac{a + b + 2x}{b + x} \bullet \text{BF} \\ \text{BH} = \frac{a - b}{a - x} \bullet \text{BF}\text{\ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\frac{a + b + 2x}{b + x} = \frac{a - b}{a - x}.\]

\[6)\ (a - x)(a + b + 2x) =\]

\[= (b + x)(a - b)\]

\[a^{2} + \text{ab} + 2\text{ax} - \text{ax} - \text{bx} - 2x^{2} =\]

\[= \text{ba} - b^{2} + \text{ax} - \text{bx}\]

\[a^{2} + 2\text{ax} - \text{ax} - 2x^{2} + b^{2} - \text{ax} =\]

\[= 0\]

\[a^{2} - 2x^{2} + b^{2} = 0\]

\[2x^{2} = a^{2} + b^{2}\]

\[x^{2} = \frac{a^{2} + b^{2}}{2}\]

\[x = \frac{\sqrt{a^{2} + b^{2}}}{2}.\]

\[Ответ:\text{MN} = \frac{\sqrt{a^{2} + b^{2}}}{2}.\]